Q. No.In the given circuit, the power developed in the \(2~\text{k}\Omega\) resistor is:
1. \(36~\text{mW}\)
2. \(12~\text{mW}\)
3. \(144~\text{mW}\)
4. \(72~\text{mW}\)
1. \(36~\text{mW}\)
2. \(12~\text{mW}\)
3. \(144~\text{mW}\)
4. \(72~\text{mW}\)
An transistor having A.C. current gain of 50 is to be used to make an amplifier of voltage gain 6. What will be the power gain in the amplifier?
1. 300
2. 600
3. 750
4. 400
Current \(I_1\) through the Zener diode shown in the circuit is:
1. Zero
2. \(0.6\) mA
3. \(0.2\) mA
4. \(0.8\) mA
| 1. | only if both inputs are zero. |
| 2. | if either or both inputs are \(1\). |
| 3. | only if both inputs are \(1\). |
| 4. | if any of the inputs is zero. |
In an NPN transistor, 108 electrons enter the emitter in s. If 1% electrons are lost in the base, the fraction of current that enters the collector and the current amplification factor are, respectively:
1. 0.7 and 50
2. 0.9 and 90
3. 0.8 and 49
4. 0.99 and 99
When a transistor is used in a common emitter mode as an amplifier:
| 1. | the base-emitter junction is forward biased. |
| 2. | the base-emitter junction is reverse biased. |
| 3. | the base-collector junction is forward biased. |
| 4. | the input is connected in series with voltage applied across the base-collector junction. |
(If diodes are ideal)
| 1. | \(0,0\) | 2. | \(5~\text{mA},5~\text{mA}\) |
| 3. | \(5~\text{mA},0\) | 4. | \(0,5~\text{mA}\) |
The LED:
| 1. | is reverse-biased. |
| 2. | is forward-biased. |
| 3. | can be made of \(\mathrm{GaAs}.\) |
| 4. | both (2) and (3) are correct. |
Logic gates \(X\) and \(Y\) have the truth tables shown below:
| \(X\) | ||
| \(P\) | \(Q\) | \(R\) |
| \(0\) | \(0\) | \(0\) |
| \(1\) | \(0\) | \(0\) |
| \(0\) | \(1\) | \(0\) |
| \(1\) | \(1\) | \(1\) |
| \(Y\) | |
| \(P\) | \(R\) |
| \(0\) | \(1\) |
| \(1\) | \(0\) |
When the output of \(X\) is connected to the input of \(Y\), the resulting combination is equivalent to a single:
1. NOT gate
2. OR gate
3. NAND gate
4. AND gate
© 2025 GoodEd Technologies Pvt. Ltd.